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Question
If `sin^-1"x" + tan^-1"x" = pi/2`, prove that `2"x"^2 + 1 = sqrt5`
Solution
`sin ^-1"x"+tan^-1"x"=pi/2`
⇒ `tan^-1("x"/ sqrt (1-"x"^2)) + tan^-1"x" =pi/2`
⇒ `tan^-1[("x"/ sqrt (1-"x"^2)+ "x")/(1 -"x"/(sqrt(1-"x"^2)). "x")] = pi/2`
⇒ `tan^-1(("x"+"x"sqrt(1-"x"^2))/(sqrt(1-"x"^2)-"x"^2)) = pi/2`
⇒ `("x"+"x"sqrt(1-"x"^2))/(sqrt(1-"x"^2)-"x"^2 )= tan(pi/2)`
⇒ `(sqrt(1-"x"^2)-"x"^2)/("x"+"x"sqrt(1-"x"^2)) = cot (pi/2)=0`
⇒ `sqrt(1-"x"^2) - "x"^2 = 0`
⇒ `"x"^2= sqrt(1-"x"^2)`
⇒ `"x"^4 = 1-"x"^2`
⇒ `"x"^4 +"x"^2 - 1 = 0`
⇒ let `"x"^2 ="t"`
⇒ t2 + t - 1 = 0
t = `(-1±sqrt(1+4))/2`
t = `(-1±sqrt(5))/2 =⇒ 2"t" = -1± sqrt5`
⇒ ` 2"x"^2+1 = ± sqrt5` ignoring negative value
We get : `2"x"^2+1 = sqrt5` = proved
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